Shell Programming and Scripting

Is there a way to figure out if given date is in DST or in EST? So imagine a shell script is_date_dst.sh

sh is_date_dst.sh 2009-03-02 would return 0 and
sh is_date_dst.sh 2009-03-22 would return 1

This script should not assume the timezone of the machine it runs on, but only the date passed in as a param.

Thanks

In the examples above how did you arrive at either 0 or 1?
In general, how can you determine the TZ by looking at a date only?

If the date is in DST (ie between 2nd Sunday of Mar and 1 sunday of November, both inclusive) then the function should return 1, else 0.

something to start with:

Code:

#!/bin/ksh
#set -x

d='2009-03-02'

year="${d%%-*}"

#echo "year->[${year}]"

dd="$(echo "${d}"|sed 's/-//g')"

march=$(printf "%d%02d%02d" "${year}" '3' $(cal 3 "${year}" | nawk 'FNR==4 {print $NF}'))
nov=$(printf "%d%02d%02d" "${year}" '11' $(cal 11 "${year}" | nawk 'FNR==3 {print $NF}'))

#echo "march->[${march}]"
#echo "nov->[${nov}]"

if [ ${dd} -gt ${march} -a ${dd} -le ${nov} ]; then
   echo 1
else
   echo 0
fi

Thanks alot for your help. This is pretty close, but I think this gets tricky if you select fields based on rows/fields (which is what you are doing). Here you are picking 2nd Saturday of Mar and 1 Saturday of Nov.
How do we then get the 2nd Sunday of Mar and 1 Sunday of Nov?

I think the most correct logic is:
for Mar: get the last day of Feb, then get the first Sunday after that date. Then add 7 to it.
for Nov: get the last day of Oct, then get the first Sunday after that date.

I dont know how this can be done in shell/AWK/sed though

Thanks.

i don't know what's DST and EST dates, but if i have to do it
I would just check the date today, check if its DST put a flag file, if it reaches EST delete the flag file. then just do whatever you want to the time subtract 1 or something.

GNU date can give you the timezone for a date...

Code:

date -d '2009-03-02' '+%Z'